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Consider a position-time graph for a particle's motion, where the y-axis is position and the x-axis is time (in seconds). A negative quartic function with an apex at (20, 10).

Now, consider the question:

At what point in time is the particle at rest?

The obvious answer is at 20 seconds, but I'm having trouble visualizing exactly why. One could say "The particle is not moving at 20 seconds, therefore it is at rest." But couldn't you say the same thing about any other point on this graph? If you froze time at 18 seconds, the particle would not be moving. "Yes, but if you unfreeze time, it is clearly moving." Again, same thing at 20 seconds. The particle isn't moving after 20 seconds, but only instantaneously. At 20.000000...1 seconds, the particle does move, so the only way to find evidence of the object's "rest" is to freeze time and watch it not move. But then again, you can freeze time at any point and the object will not move. So what is rest? The absence of movement? Or just the absence of velocity?

Apologies if this seems like an extremely stupid question.

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    $\begingroup$ Sure you could freeze time but that's an artificial condition to make the object appear to be motionless, it still technically has a nonzero instantaneous velocity at that time, thus it is in motion and not at rest. The question is all about instanteous velocity here. $\endgroup$
    – Triatticus
    Mar 23, 2022 at 19:10
  • $\begingroup$ But what exactly are motion and rest? $\endgroup$ Mar 23, 2022 at 19:11
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    $\begingroup$ As I said it's a condition of instantaneous velocity, that's all. If an object has nonzero instantaneous velocity it is clearly in motion, if it is zero it is at rest $\endgroup$
    – Triatticus
    Mar 23, 2022 at 19:13
  • $\begingroup$ I could take a photo of a F1 car moving at 300 kmph and the car would look motionless... would that be considered as a conclusive evience that the car was at rest when the photo was taken? $\endgroup$ Mar 24, 2022 at 8:37
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    $\begingroup$ This is the exact line of logic used in Zeno of Elea's Arrow Paradox. $\endgroup$ Mar 31, 2022 at 22:37

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Dissenting opinion: I would not say that the particle is ever "at rest." That plot looks roughly like the answer to, "what is the height of a ball that is thrown straight up into the air?"

OK, Suppose we throw a ball straight up into the air. After the initial impulse, it's velocity is upward, but it is accelerating downward. There comes a moment, the peak of its flight, when the velocity changes from upward to downward. For an instant, the ball's velocity is zero (t=20 in your plot.) But at that moment, it's still accelerating downward.

Would you really say that the ball is "at rest?" if only for that infinitesimal instant during its flight?

I would not. I don't find the phrase "at rest" to be a good description of accelerated motion.


What exactly is rest?

It's a colloquialism. It isn't formal physics language. If we want to be formal, we talk about position, and velocity, and acceleration, etc. We don't use vague words like "rest."

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I am answering this in context of introductory physics (so that there are no issues of reference frames).

To be "at rest" means to have zero velocity, which can be visualized on a position-vs-time graph (as you have here) when the tangent line [to the worldline] has zero slope (which, of course, can be expressed by saying when the time-derivative of the position is zero).

From the data on the graph, this only occurs at $t=20\rm{\ s}$... and at no other instant. (The slope of the tangent line at $t=18\rm{\ s}$ is positive (and is thus nonzero).)

[Instantaneous] Velocity is a rate-of-change of position with respect to time. It requires information from at least two instants (two events) in order to form (or approximate) a slope.

Your act of "Freezing" [by effectively looking at the position at only one instant] only tells you where you are at that instant, but not what your rate-of-change is. (Your act of "Freezing" does not make the slope at $t=18\rm{\ s}$ zero.)

You need the slope at that instant (and not the position) ... or you need a nearby (possibly infinitesimally nearby) second event's position and time with your first event's position and time.

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In Newtonian mechanics, the state of a system at a time is represented by the zeroth and first derivatives of position (i.e., by position and velocity). The equations of motion give you the second derivative (acceleration) in terms of the zeroth and first.

So it doesn't really make sense to freeze time in the way you've done it. If you want the system to keep evolving as though nothing had happened when you unfreeze it, then the frozen state needs to include the velocities.

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In the context of the first Newton’s law of motion the better term would be equilibrium

If there are no forces acting on the body, or the net force is equal to zero, the object is in equilibrium.

From this we can conclude that equilibrium means that the object acceleration is zero, which indicates that the object is moving at constant velocity (which can be zero, but does not have to) in a straight line.

Note that you can always find an inertial reference frame in which object velocity is zero. That is why equilibrium is much better term than rest.

so the only way to find evidence of the object's "rest" is to freeze time and watch it not move.

Object velocity at certain point in time is defined as tangent of its path (displacement) at that time instant. Object velocity is zero when the tangent slope is zero. You do not have to freeze time, just find the tangent.

In your example the object velocity is zero at $t = 20 \text{ s}$. If you define the "rest" as "zero velocity", than the object is at rest at $t = 20 \text{ s}$. When the object is at rest only at some particular time instant, e.g. when velocity switches from negative to positive and vice-versa, we say that the object is "instantaneously at rest".

If you froze time at 18 seconds, the particle would not be moving.

In real world, that (unfortunately) you cannot do.

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  • $\begingroup$ But can equilibrium really be synonymous with rest? If you push a mass on a frictionless surface, the object will continue to move forever with constant velocity. In this case the object is at equilibrium because of its constant velocity, but not at rest -- it's clearly moving. $\endgroup$ Mar 23, 2022 at 19:03
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    $\begingroup$ This answer doesn’t address the graph in the question, where at 20 there is (potentially) a net force on the object but no instantaneous motion. $\endgroup$ Mar 23, 2022 at 19:05
  • $\begingroup$ @FouadSaffar Exactly for the reason you mention I used “equilibrium” rather than “rest”. In your example, can you find an inertial reference frame in which the object velocity is zero? Sure, you can say the object is at rest when its velocity is zero. For objects in equilibrium you can always find inertial reference frame in which velocity will be zero. $\endgroup$ Mar 23, 2022 at 19:16
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Rest is zero velocity with respect to the inertial frame of reference chosen, which must remain consistent throughout this analysis. As was mentioned in a comment, this occurs when the tangent (first derivative with respect to time) of the position-time graph is zero.

This is separate from equilibrium, in that there are still net forces being applied (thus the acceleration "down"), but at this moment in this inertial frame of reference, the velocity is zero.

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Motion is a change of position over a period of time. If you focus on a very short interval of time on the graph around the 18s mark, the change in the position of the particle will be correspondingly small, but it will not be zero. You can make the time period as close to zero as you like, and the change in position will still not be zero- specifically the ratio of the change in position to the change in time will always be equal to the speed of the particle. It is only at 20s that the speed of the particle will be zero, and hence that the change in position over an infinitesimally short period of time will also be zero.

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At time $ 18 s $ you can expand for the motion for small times different from $ t = 18 s $ as $$ s ( t ) = 0 + \frac{ d s }{ d t } ( t - 18 ) + \frac{ 1 }{ 2 } \frac{ d^2 s }{ d t^2 } ( t - 18 )^2 + \cdots $$ where the derivatives are taken at $ ( s,t ) = ( 0, 18 ) $ and is just the Taylor expansion at that point. Note that for $ t = 18 s, d s/ dt \ne 0 $ so the particle is not stationary.

At time $ 20 s $ you have $$ s ( t ) = 10 + \frac{ d s }{ d t } ( t - 20 ) + \frac{ 1 }{ 2 } \frac{ d^2 s }{ d t^2 } ( t - 20 )^2 + \cdots $$ where the derivatives are now taken at $ t = 20 s $. But at $ t = 20 s $ you have $ d s/d t = 0 $, so that the particle is momentarily at rest. However, the acceleration term, $ d^2 s/d t^2,$ is non-zero, so that the particle at this point will change its speed.

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